zoj2750

Idiomatic Phrases Game

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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Tom is playing a game called Idiomatic Phrases Game. An idiom consists of several Chinese characters and has a certain meaning. This game will give Tom two idioms. He should build a list of idioms and the list starts and ends with the two given idioms. For every two adjacent idioms, the last Chinese character of the former idiom should be the same as the first character of the latter one. For each time, Tom has a dictionary that he must pick idioms from and each idiom in the dictionary has a value indicates how long Tom will take to find the next proper idiom in the final list. Now you are asked to write a program to compute the shortest time Tom will take by giving you the idiom dictionary.

Input

The input consists of several test cases. Each test case contains an idiom dictionary. The dictionary is started by an integer N (0 < N < 1000) in one line. The following is N lines. Each line contains an integer T (the time Tom will take to work out) and an idiom. One idiom consists of several Chinese characters (at least 3) and one Chinese character consists of four hex digit (i.e., 0 to 9 and A to F). Note that the first and last idioms in the dictionary are the source and target idioms in the game. The input ends up with a case that N = 0. Do not process this case.

Output

One line for each case. Output an integer indicating the shortest time Tome will take. If the list can not be built, please output -1.

Sample Input

5
5 12345978ABCD2341
5 23415608ACBD3412
7 34125678AEFD4123
15 23415673ACC34123
4 41235673FBCD2156
2
20 12345678ABCD
30 DCBF5432167D
0

Sample Output

17
-1

 

#include <cstdio>
#include <cstring>
#define INF 1000000000    //无穷大
#define MAXN 1000        //顶点个数最大值

struct idiom
{
    char front[5], back[5];    //存储成语第1个和最后一个汉字
    int T;    //选用这个成语后，Tom需要花时间T才能找到下一个合适的成语
};

idiom dic[MAXN];        //字典
int Edge[MAXN][MAXN];    //邻接矩阵
int dist[MAXN];    //求得的源点0到每个顶点的最短路径长度
int S[MAXN];    //S[i]=1表示顶点i的最短路径已求得
int N;            //成语个数

int main( )
{
    int i, j, k;//循环变量
    char s[100];//读入的每个成语(取其第1个和最后一个汉字)
    int len;    //成语的长度
    while( scanf( "%d", &N ) != EOF )
    {
        if( N==0 )  break;    //输入结束
        for( k=0; k<N; k++ )
        {
            scanf( "%d%s", &dic[k].T, s );
            len = strlen(s);
            for( i = 0,j = len-1; i < 4; i++, j-- )    //取前4个字符、后4个字符
            {
                dic[k].front[i] = s[i];
                dic[k].back[3-i] = s[j];
            }
            dic[k].front[4] = dic[k].back[4] = '\0';
        }
        
        for( i=0; i<N; i++ )    //建图
        {
            for( j=0; j<N; j++ )
            {
                Edge[i][j] = INF;
                if( i==j )  continue;
                if( strcmp( dic[i].back, dic[j].front ) == 0 )
                    Edge[i][j] = dic[i].T;
            }
        }
        //Dijkstra算法
        for( i=0; i<N; i++ )    //初始化
        {
            dist[i] = Edge[0][i];  S[i] = 0;
        }
        S[0] = 1;  dist[0] = 0;    //顶点0加入到顶点集合S
        for( i=0; i<N-1; i++ )    //确定N-1条最短路径(Dijkstra算法)
        {
            int min = INF, u = 0;
            
            //选择当前集合T中具有最短路径的顶点u
            for( j=0; j<N; j++ )
            {
                if( !S[j] && dist[j]<min )
                {
                    u=j;  min = dist[j];
                }
            }
            
            S[u] = 1;    //将顶点u加入到集合S，表示它的最短路径已求得
            
            for( k=0; k<N; k++ )    //修改T集合中顶点的dist数组元素值
            {
                if( !S[k] && Edge[u][k]<INF && dist[u] + Edge[u][k] < dist[k] )
                    dist[k] = dist[u] + Edge[u][k];
            }
        }
        if( dist[N-1]==INF )  printf( "-1\n" );
        else  printf( "%d\n", dist[N-1] );
    }
    return 0;
}